[SOLVED] Calculate specifications for Cockcroft-Walton generator

[Blasterg]

Good night, had a few questions about the design of a high voltage power supply...

I'm currently in the development stage of a power supply for a tube of X-rays, the initial voltage will come out from a source of 24V 10A DC, which fed a ZVS driver for a flyback, which will output approximately 24kV, which is not a negligible amount, and my tube (CEI OX / 70-G4) will work fine, but I would add a CW multiplier to achieve an output of at least 65-70kV (since the tube has a rated voltage 75KV), could help me to calculate the number of phases and specifications that should have the diodes and capacitors please? : /

By my calculations, 2 phases suffice to raise the voltage from 24kV to 67.88kV, but I'm a little confused with the diodes and capacitors that I should use.

By the way, I still do not have in my hands the power supply to deliver the initial voltage, it's a Meanwell S-250-24
I noticed that comes with a potentiometer to adjust the output voltage, and I was thinking maybe I could take him down to 15V, is this possible?
Because in that case I would use these 15V to have an output of approximately 15kV on the flyback and then use a CW multiplier of 4 phases and obtain 84.85kV, which would come to be 75-78kV for losses.
Help please!

BTW, sorry for my english, my native language is spanish.

 

[chuckey]

have you looked here :- https://en.wikipedia.org/wiki/Cockcroft–Walton_generator In general the capacitors should be big then getting smaller to keep the voltage pulse constant despite the current reducing up the network. I wonder what sort of high frequency extra high voltage diode you are going to use. Old valve colour TV sets used a solid state tripler to get their 24 KV (at 16 KHZ), could be the cheapest way to go?
Frank

 

[BradtheRad]

With the CW multiplier...

Each capacitor acquires a charge equal to the supply voltage.

Each diode must be rated to withstand the supply voltage.

Here is a simulation showing (theoretically) how 24 kV AC square waves might be fed to a CW tripler. I believe you'll need 3 stages (if you want upwards of 70 kV).



Do you have 24 kV as a sine wave? Then its peaks are 33.6 kV. In that case you could get by with just two multiplier stages, to get 66 kV DC output. That is only if your load is light.

If your load is substantial, then you must increase the value of all capacitors.

 

[Blasterg]

Thanks for your replies!
I'll use this ZVS driver+flyback:
https://www.ebay.com/itm/111426038219

"12-36VDC input (recommend 24V), output high voltage direct current, voltage of the input voltage of about 1000 times. The 12V input can reach 50-100W power, 24 V input max can reach more than 200W, so request power supply current best in more than 10A."
How can I know the output frequency of that driver?
Ohh and the output voltage is DC, not AC, anyway I read that these multiplier can work with both (AC and pulsing DC), so...
I think that an CW multiplier of 2-stages can work nice, I'll settle with 65kV, don't want to put to work the tube at full power. (75kV) I just want better contrast on the resultant image with my intensifier screen when placing a denser object.
These pieces will work then?
- 30KV 5mA High Voltage Diode
- 30KV 1000PF High Voltage Ceramic Capacitors

 

[BradtheRad]

If you use 5mA diodes, then you must limit incoming Amperes to 5 mA. This means that upstream somewhere there needs to be sufficient impedance (several megohms, according to my simulation).

A resistor inline will do this. The problem is that it reduces final output voltage. Therefore you would bypass it at a strategic time, so output will reach maximum voltage.

Ignore the above if your 24kV supply is limited to 5 mA.

 

[Blasterg]

If you use 5mA diodes, then you must limit incoming Amperes to 5 mA. This means that upstream somewhere there needs to be sufficient impedance (several megohms, according to my simulation).

A resistor inline will do this. The problem is that it reduces final output voltage. Therefore you would bypass it at a strategic time, so output will reach maximum voltage.

Ignore the above if your 24kV supply is limited to 5 mA.

:O
In that case...
I'll use 10mA diodes to get the maximum power from my power supply

I already figured all my possible currents, trying to avoid surpass 240W and based on x-ray tube specifications:
24kV:
2.5V 2.0A Filament + 24kV 3mA Anode = 72W = 3.6W X-Ray [Based on 5% efficiency of x-rays produced]
3.0V 2.3A Filament + 24kV 5mA Anode = 120W = 6W X-Ray
3.5V 2.5A Filament + 24kV 7mA Anode = 168W = 8.4W X-Ray
4.0V 2.7A Filament + 24kV 8mA Anode = 192W = 9.6W X-Ray

65kV: (with 2-stages half-wave CW generator)
2.5V 2.0A Filament + 65kV 3mA Anode = 195W = 9.75W X-Ray
2.75V 2.2A Filament + 65kV 4mA Anode = 260W = 13W X-Ray [Overload, but within acceptable ranges]

This is my final machine, please correct me if it's wrong

Sorry for the graphics, I made it entirely on Paint. xD

 

[BradtheRad]

Your schematic seems to indicate that the 24 kV DC will be chopped into DC pulses, and fed to a single-polarity Villard multiplier. The result will be lower output than when you feed AC to a symmetrical multiplier.

I think you'll find you need to add more stages of Villard cells.

Also there is the question whether your X-ray emitter should have a negative supply? This affects how you design your flyback and voltage multiplier.

 

[Blasterg]

Your schematic seems to indicate that the 24 kV DC will be chopped into DC pulses, and fed to a single-polarity Villard multiplier. The result will be lower output than when you feed AC to a symmetrical multiplier.

I think you'll find you need to add more stages of Villard cells.

Also there is the question whether your X-ray emitter should have a negative supply? This affects how you design your flyback and voltage multiplier.

How many stages you recommend to get, ummmm 65-70kV?
Well, I've read that some people did not connect the negative supply in this same tube and it worked fine, so I'll leave it disconnected.

 

[FvM]

I wonder what kind of switch you imagine for your circuit? I'm not aware of a suitable device by state-of-the-art.

Secondly which "negative supply" should be disconnected?

 

[BradtheRad]

How many stages you recommend to get, ummmm 65-70kV?

My simulation has 5 Villard cells. (I suppose you would call it 2.5 stages). 24 kV DC pulses input.
Output is 65 kV at 3 mA.
200W.



The flyback needs to provide much more than 200W. The diodes particularly will need higher ratings than you have mentioned.
Not to mention hundreds of mA which the multiplier will try to draw after power-up.

It may be easier to take your flyback's secondary output, and apply it directly to a symmetrical CW multiplier. This would let you get by without a switching device made for high voltage.

Although the flyback transformer does not output symmetrical AC, it may still be satisfactory to use it in the fashion of my post #3 simulation.

 

[Blasterg]

I wonder what kind of switch you imagine for your circuit? I'm not aware of a suitable device by state-of-the-art.

Secondly which "negative supply" should be disconnected?

I know, no switch (small and cheap) can cut securely these voltages, was only added illustratively.
The negative pole coming out of the flyback, it's optional for this tube, I'll just heat up the filament.

My simulation has 5 Villard cells. (I suppose you would call it 2.5 stages). 24 kV DC pulses input.
Output is 65 kV at 3 mA.
200W.



The flyback needs to provide much more than 200W. The diodes particularly will need higher ratings than you have mentioned.
Not to mention hundreds of mA which the multiplier will try to draw after power-up.

It may be easier to take your flyback's secondary output, and apply it directly to a symmetrical CW multiplier. This would let you get by without a switching device made for high voltage.

Although the flyback transformer does not output symmetrical AC, it may still be satisfactory to use it in the fashion of my post #3 simulation.

Umm, I'll take the second option, then the schematic would be like this?

What diodes you recommend?
And that 100k resistor is necessary?

 

[BradtheRad]

Umm, I'll take the second option, then the schematic would be like this?

Yes, that is what I was thinking. No guarantee it will succeed. It's all theoretical at this stage, of course. Just trying to think up a more straightforward topology.

What diodes you recommend?

This is unknown territory for me. I don't know whether individual diodes are made for 30 or 40 kV, or whether you'll need to string together several diodes with a lesser rating.

And that 100k resistor is necessary?

No, it's purpose is to mimic the presence of some amount of input impedance. It's useful to observe what happens when something limits Ampere draw. (Example, so that no diode is exposed to more than 10mA.)

Any amount of input resistance drops voltage, so as a result it would prevent the final output voltage from rising as high as it could.

So you shouldn't need to install a resistor, except in some peculiar circumstance where it turns out that you really do need to limit current.

 

[chuckey]

Instead of current limiting the output, limit the input current/voltage on the converter, its easier working at 24V rather then 60 KV
Frank

 

[Blasterg]

Instead of current limiting the output, limit the input current/voltage on the converter, its easier working at 24V rather then 60 KV
Frank

I know, but x-ray tubes need high voltages to get images with more contrast, because the rays have low power of penetration with voltages lower than 50kV. :|
And sure, I'll NOT use this machine to get radiographs of people, just to see through things like motherboards and other PC stuff, all this using the required lead protection and safer distances.

Yes, that is what I was thinking. No guarantee it will succeed. It's all theoretical at this stage, of course. Just trying to think up a more straightforward topology.



This is unknown territory for me. I don't know whether individual diodes are made for 30 or 40 kV, or whether you'll need to string together several diodes with a lesser rating.



No, it's purpose is to mimic the presence of some amount of input impedance. It's useful to observe what happens when something limits Ampere draw. (Example, so that no diode is exposed to more than 10mA.)

Any amount of input resistance drops voltage, so as a result it would prevent the final output voltage from rising as high as it could.

So you shouldn't need to install a resistor, except in some peculiar circumstance where it turns out that you really do need to limit current.

Thank you so much! :thumbsup: I'll try when receive the components, and will leave it with 30kV 10mA diodes, what can go wrong? xD
Anyway, it's only an experiment, if something burns at least I'll have learned.